The big idea: Most elements exist as a mixture of isotopes — atoms of the same element with different masses (different numbers of neutrons).
So the relative atomic mass, Ar, quoted on the periodic table is not the mass of a single atom. It is the weighted average mass of all the isotopes, taking into account how common each one is.
Ar has no units — it is a ratio, comparing the average atom's mass with one-twelfth of a carbon-12 atom.
Mass spectrum of chlorine: two isotope peaks — ³⁵Cl (75%) and ³⁷Cl (25%). The taller peak is the more abundant isotope.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Why chlorine's A_{r} is 35.5: Chlorine is 75% ³⁵Cl and 25% ³⁷Cl. No chlorine atom actually has a mass of 35.5 — that value is just the average, pulled close to 35 because the lighter isotope is three times more common.
A mass spectrometer sorts the atoms of a sample by mass. It gives a mass spectrum: a bar chart with the mass-to-charge ratio (m/z) on the x-axis and the relative abundance of each isotope on the y-axis.
Reading the chart: For singly-charged ions the m/z value is the isotope's mass. The height of each peak gives the relative abundance.
- The number of peaks = the number of isotopes. - The position of a peak (m/z) = the mass of that isotope. - The height of a peak = how abundant that isotope is.
Mass spectrum of magnesium: three isotopes — ²⁴Mg (79%), ²⁵Mg (10%), ²⁶Mg (11%). Read each m/z value and its relative abundance.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Abundances may not be given as %: Sometimes the heights are given as relative values (e.g. 7.9 : 1.0 : 1.1) instead of percentages. Convert to percentages by dividing each by the total and multiplying by 100 — or just divide the weighted sum by the total of the abundances instead of by 100.
Learn what examiners really want
See exactly what to write to score full marks. Our AI shows you model answers and the key phrases examiners look for.
To find Ar, multiply each isotope mass by its percentage abundance, add the results, then divide by 100. Each isotope is weighted by how common it is.
- relative atomic mass (no units — it is a ratio)
- the m/z value of each isotope peak
- the relative abundance of that isotope, as a percentage
- divides by the total percentage, giving a weighted mean
Worked example — Ar of chlorine (two isotopes)
Chlorine has two isotopes: ³⁵Cl with 75.0% abundance and ³⁷Cl with 25.0% abundance. Calculate the relative atomic mass of chlorine.
Solution
- Formula first — weight each isotope mass by its abundance:
- Substitute the two isotopes:
- Work out the top line:
- Divide by 100:
Final answer
Ar(Cl) = 35.5 (no units) — closer to 35 because the lighter isotope is more abundant.
Worked example — Ar of magnesium (three isotopes)
Magnesium has three isotopes: ²⁴Mg (78.6%), ²⁵Mg (10.1%) and ²⁶Mg (11.3%). Calculate the relative atomic mass of magnesium to three significant figures.
Solution
- Formula first — extend the weighted sum to all three isotopes:
- Substitute each isotope:
- Work out the top line:
- Divide and round to 3 s.f.:
Final answer
Ar(Mg) = 24.3 — dominated by the most abundant isotope, ²⁴Mg.
How this is tested: This skill turns up as a short Paper 2 calculation ('calculate the relative atomic mass from the two isotope abundances') and as a Paper 1A MCQ that asks you to read peaks off a mass spectrum.
You are given the isotope masses and their abundances (as a chart or in words) and must return the weighted-average A_{r} — usually to a stated number of significant figures.
Where the marks go: Mark 1 is the correct set-up — masses multiplied by abundances and divided by the total. Mark 2 is the correct final value to the right significant figures. Always sanity-check: Ar must lie between the lightest and heaviest isotope, closest to the most abundant one.
IB-style question — Ar of element X
A sample of element X gives two peaks in its mass spectrum: m/z = 63 at 69.2% abundance and m/z = 65 at 30.8% abundance. Calculate the relative atomic mass of X to one decimal place. [2]
Solution
- Formula first — weight each isotope mass by its abundance:
- Read the two peaks off the spectrum and substitute:
- Work out the top line:
- Divide and round to 1 d.p.:
Final answer
Ar(X) = 63.6 (the element is copper — closer to 63 because ⁶³Cu is the more abundant isotope).